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sergeinik [125]

4 years ago

6

Is it true that atmospheric oxygen gas is a pure substance?

1 answer:

Mariana [72]4 years ago

3 0

The stronger the forces, the more rigid the matter. Pure Substances are made of the same material throughout and have the same properties throughout. Pure substances cannot be separated into other substances. Some examples are carbon, iron, water, sugar, salt, nitrogen gas, and oxygen gas.

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Why would the bulb not light?

Norma-Jean [14]

Im not sure this is correct but, i think thats because the wire connected at the negative side. The current go from positive to negative and thats why the bulb not light, also the bulb location is not at the center of the battery which explain why there's no current at the upper side.

Hope you understand. Sorry if my english sucks YEET

6 0

3 years ago

Read 2 more answers

Hallar la energía mecánica de un pendulo , en un punto de su oscilación donde la energía cinética es 0,39 J y en ese mismo punto

iris [78.8K]

Answer:

E = 0,39 + 0,59 = 0,98 J

Explanation:

7 0

3 years ago

How does energy relate to force?

ioda

Answer:work is force times distance

Explanation:to create a force u need energy and the greater the energy the greater the force is applied to an object.

4 0

3 years ago

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

siniylev [52]

Answer:

r = 1.86 m

Explanation:

Here the force due to wall of the cylinder is towards the axis of the cylinder

This force will act as centripetal force for the people sit inside the chamber

now we will have

$F_c = \frac{mv^2}{R}$

now we will have

$F_c = 488 N$

m = 84.4 kg

v = 3.28 m/s

now we have

$488 = \frac{84.4(3.28)^2}{r}$

now we have

$r = 1.86 m$

8 0

3 years ago

An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?

dimulka [17.4K]

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

$V/V_g = V - V_g = 5.2 - 0 = 5.2 \ m/s$

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

5 0

3 years ago